Help with Greens function/Fourier transformation to solve screened Poisson equation I am having trouble getting from one line to the next from this wiki page.  I am referring to the text line 

Green's function in $r$ is therefore given by the inverse Fourier transform, 

where

$$G(r) ~=~ \frac{1}{(2\pi)^3} \iiint d^3k \frac{e^{i{\bf k}\cdot{\bf  r}}}{k^2+\lambda^2}$$

goes to

$$G(r) ~=~ \frac{1}{2\pi^2r} \int^{\infty}_0 \!dk_r \frac{k_r \sin(k_r r)}{k_r^2+\lambda^2}.$$

Where does the $\frac{1}{r}$ term come from and what is $k_r$? How did they simplify the triple integral? Divergence theorem? Stokes? Detailed steps would be much appreciated.
 A: You convert the integral to spherical coordinates and the $k \cdot r$ term becomes $kr\cos\theta$.  Integrating over $\theta$ and $\phi$ gets you the single dimensional integral expression.  
A: I ran into the same problem today and didn't like the use of the Bessel functions. So here's my simpler approach to it: no scary stuff.
Put everything into spherical coordinates: 
$k_1=k \sin{\theta} \cos{\phi}, k_2=k \sin{\theta} \sin{\phi}, k_3=k \cos{\theta}$, and $d^3 k =k^2 \sin{\theta} dkd \theta d \phi$ and $r_1=r \sin{\theta}' \cos{\phi}', r_2=r \sin{\theta}' \sin{\phi}', r_3=r \cos{\theta}'.$
Then you make the dot product and it gives $k\cdot r \cdot [...]$, where $[...]$ is an ugly term full of θ,θ',ϕ and ϕ' for the angle between k and r.
However, you can choose a different orthogonal axis: say one where z' in parallel to r. In these "r-ortogonal" axes: $\hat x_r$' = $\hat x_r$ ^ $\hat z_r$' and $\hat y_r$' = $\hat z_r$' ^ $\hat x_r$' and $\hat z_r$' = r/r.
So now your spherical coordinates are:
$k_1=k_r\sin{\theta}\cos{\phi}$, $k_2=k_r\sin{\theta}\sin{\phi}$, $k_3=k_r\cos{\theta}$, and $d^3k =k_r^2\sin{\theta}dkd{\theta}d{\phi}$ and $r_1=0, r_2=0, r_3=r$.
and the dot product is simply $k_r\cdot\ r\cdot \cos{\theta}$, with $\theta$ the angle between k and r in these "r-axes".
Now define $\gamma = k_r\cdot\ r\cdot \cos{\theta}$ and dγ = $k_r\cdot\ r\cdot \sin{\theta}d{\theta}$ and substitute in the integrals. This will give you the $k_r/r$ term. 
When you integrate the $e^{i\gamma}$ from $\gamma = -k_rr$ to +$k_rr$, this will wive you the $\sin{k_rr}$.
